Show that an ideal is proper if and only if it does not contain 1.

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- Jul 27th 2009, 01:11 PMpeteryellowring
Show that an ideal is proper if and only if it does not contain 1.

- Jul 27th 2009, 03:01 PMsiclar
Prove these two statements: If 1 is in the ideal, it is not proper. If 1 is not in the ideal, it is proper. The second statement is obvious by the definition of being a proper ideal, i.e. if the ring is and ideal is , since .

Use the definition of an (left) ideal: for all elements of the ring, . In particular, if then for any . What if ? - Jul 27th 2009, 09:07 PMGamma
Does your ring definition require all rings to be unitary?

If you take the ring to be . Certainly and it does not contain 1, but is not a proper ideal since it is the whole ring.